The transfer function of a phase lead compensator is given by D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )} The frequency (in rad/sec), at which \angle D(j\omega ) is maximum, is
T(s)=\frac{1+3Ts}{1+Ts} Frequency at which \angle T(j\omega ) is maximum, (\omega _m)=\frac{1}{T\sqrt{\alpha }} \alpha =\frac{1}{1/3}=3 \omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}
Question 2
The transfer function C(s) of a compensator is given below. C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})} The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
C_1=\frac{10(s+1)}{(s+10)} Zero at s=-1 Pole at s=-10 As zero is closer origin, zero dominates pole. Hence C_1 is lead compensator. C_2=\frac{s+10}{10(s+1)} Zero at s=-10 Pole at s=-1 As pole is closer to origin, pole dominates zero. Hence C_2 is lag compensator.
Question 7
The system \frac{900}{s(s+1)(s+9)} is to be compensated such that its gain-crossover frequency
becomes same as its uncompensated phase crossover frequency and provides a 45^{\circ} phase margin. To achieve this, one may use
A
a lag compensator that provides an attenuation of 20 dB and a phase lag of 45^{\circ} at the frequency of 3 \sqrt{3} rad/s
B
a lead compensator that provides an amplification of 20 dB and a phase lead of 45^{\circ} at the frequency of 3 rad/s
C
a lag-lead compensator that provides an attenuation of 20 dB and phase
lag of 45^{\circ} at the frequency of \sqrt{3} rad/s
D
a lag-lead compensator that provides an attenuation of 20 dB and phase
lead of 45^{\circ} at the frequency of 3 rad/s
Let uncompensated syatem, T(s)=\frac{900}{s(s+1)(s+9)} Phase crossover frequency of uncompensated system =(\omega _{pc}), at this frequency phase of T(j \omega ) is -180^{\circ} Put s=j\omega \text{ in } T(s) \begin{aligned} T(j\omega )&=\frac{900}{j\omega (j\omega +1)(j\omega +9)} \\ \angle T(j\omega )&=-90^{\circ}-tan ^{-1}\omega -tan^{-1}\left ( \frac{\omega }{9} \right ) \\ \text{At, } & (\omega _{pc})_1,\angle T(j\omega )=-180^{\circ}\\ -180^{\circ}&=-90^{\circ}-tan^{-1}(\omega _{pc})_1 -tan^{-1}\left ( \frac{(\omega _{pc})_1}{9} \right )\\ -90^{\circ}&=tan^{-1}\frac{(\omega _{pc})_1+\frac{(\omega _{pc})_1}{9}}{1-\frac{(\omega _{pc})^2_1}{9}} \\ \Rightarrow 1-\frac{(\omega _{pc})^2_1}{9}&=0 \\ \Rightarrow (\omega _{pc})_1&= 3 \text{ rad/sec} \end{aligned} Gain cross frequency of compensated system, (\omega _{gc})_2= phase cross frequency of uncompensated system, (\omega _{pc})_1 \Rightarrow \;\;(\omega _{gc})_2=(\omega _{gc})_1=\; 3 rad/sec Phase-margin =180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{gc})_2} \Rightarrow \; \; \; 45^{\circ}=180^{\circ}+\angle T(j\omega )|_{\omega =(\omega _{pc})_2} At (\omega _{gc})_2=3 rad/sec, phase angle of \angle T(j\omega ) is -135^{\circ} and phase of uncompensated systen is -180^{\circ} at 3 rad/sec. Therefore, the compensator provides phase lead of 45^{\circ} at the frequency of 3 rad/sec. Let X dB is the gain provided by the compensator, so at gain cross frequency, |T(j\omega )|_{com}=1 or 0 dB. Gain of uncompensated system =|T(j\omega )|_{un-com} =\frac{100}{\omega \sqrt{1+\omega ^2}\sqrt{1+\left ( \frac{\omega }{9} \right )^2}} |T(j\omega )|_{un-com} in dB \;=40-20 \log \omega -20 \log \sqrt{1+\omega ^2} -20 \log \sqrt{1+\left ( \frac{\omega }{9} \right )^2} Gain of compensated system, |T(j\omega )|_{com}=X+|T(j\omega )_{un-com}| |T(j\omega )|_{com} must be zero at gain cross frequency (\omega _{gc})_2 |T(j\omega_{gc} )_2|_{com}=X+40-20 \log (\omega _{gc})_2 -20 \log \sqrt{1+(\omega _{gc})^2_2}-20 \log \sqrt{1+\frac{(\omega _{gc})^2}{9^2}}=0 X+40-20 \log 3-20 \log \sqrt{1+3^2}-20 \log \sqrt{1+\left ( \frac{3}{9} \right )^2}=0 X=-20dB So, the compensator provides an attenuation od 20dB. Hence option (D) is correct.
Question 8
A lead compensator used for a closed loop controller has the following transfer
function \frac{K(1+\frac{s}{a})}{1+\frac{s}{b}} For such a lead compensator
Transfer function=\frac{k\left ( 1+\frac{s}{a} \right )}{\left ( 1+\frac{s}{b} \right )} Zero of TF=-a Pole of TF=-b For a lead-compensator, the zero is nearer to origin as compared to pole, hence the effect of zero is dominant, therefore, the lead-compensator when introduced in series with forward path of the trnasfer function the phase shift is increased. So, from pole-zero configuration of the compensator a \lt b.